What are the solutions to the equation 2x^2=3x+9 ?

The solutions to the equation 2x^2=3x+9 are x=3,x=-3/2

Find the zeros of 2x^2=3x+9

The zeros of 2x^2=3x+9 are x=3,x=-3/2

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2 x^{2} = 3 x + 9

x = 3, x = - \frac{3}{2}

Decimal Notation

x = 3, x = - 1.5

Solution steps

2 x^{2} = 3 x + 9

Move

9

to the left side

2 x^{2} - 9 = 3 x

Move

3 x

to the left side

2 x^{2} - 9 - 3 x = 0

Write in the standard form

a x^{2} + b x + c = 0

2 x^{2} - 3 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 9 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 9) = 9

x_{1, 2} = \frac{- ( - 3 ) \pm 9}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 9}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 9}{2 \cdot 2}

x = \frac{- ( - 3 ) + 9}{2 \cdot 2} : 3

x = \frac{- ( - 3 ) - 9}{2 \cdot 2} : - \frac{3}{2}

The solutions to the quadratic equation are:

x = 3, x = - \frac{3}{2}

\frac{18 x ^{2} y ^{3} z ^{2}}{30 x ^{2} y z ^{4}}

x^{4} y^{4}

q^{2} = - 25

\frac{3}{3} \cdot \frac{3}{3}

\frac{( x + 2 ) ( - x + 2 )}{2 x + 4} \leq 0

What are the solutions to the equation 2x^2=3x+9 ?
The solutions to the equation 2x^2=3x+9 are x=3,x=-3/2
Find the zeros of 2x^2=3x+9
The zeros of 2x^2=3x+9 are x=3,x=-3/2